CBSE Class 12 Mathematics Chapter 7: Integrals NCERT Solutions

We need to find a function whose derivative is $\sin 2x$. We know that the derivative of $\cos x$ is $-\sin x$. Let's consider the function $\cos 2x$.

The derivative of $\cos 2x$ with respect to $x$ is:

$\backslash frac\{d\}\{dx\}(\backslash cos\; 2x)\; =\; -\backslash sin\; 2x\; \backslash times\; \backslash frac\{d\}\{dx\}(2x)\; =\; -2\backslash sin\; 2x$

We want to find a function whose derivative is $\sin 2x$. From the above, we can see that:

$$\backslash sin\; 2x\; =\; -\backslash frac\{1\}\{2\}\; \backslash frac\{d\}\{dx\}\; (\backslash cos\; 2x)$$

Using the property of derivatives that $\frac{d}{dx}(k \cdot f(x)) = k \cdot \frac{d}{dx}(f(x))$, we can rewrite this as:

$$\backslash sin\; 2x\; =\; \backslash frac\{d\}\{dx\}\; \backslash left(\; -\backslash frac\{1\}\{2\}\; \backslash cos\; 2x\; \backslash right)$$

Therefore, the anti-derivative of $\sin 2x$ is $-\frac{1}{2}\cos 2x$.

Answer: The anti-derivative of $\sin 2x$ is $-\backslash frac\{1\}\{2\}\backslash cos\; 2x$.

We are looking for a function whose derivative is $\cos 3x$. We know that the derivative of $\sin x$ is $\cos x$. Let's consider the function $\sin 3x$.

The derivative of $\sin 3x$ with respect to $x$ is:

$\backslash frac\{d\}\{dx\}(\backslash sin\; 3x)\; =\; \backslash cos\; 3x\; \backslash times\; \backslash frac\{d\}\{dx\}(3x)\; =\; 3\backslash cos\; 3x$

We want to find a function whose derivative is $\cos 3x$. From the above, we can write:

$$\backslash cos\; 3x\; =\; \backslash frac\{1\}\{3\}\; \backslash frac\{d\}\{dx\}\; (\backslash sin\; 3x)$$

Using the property of derivatives that $\frac{d}{dx}(k \cdot f(x)) = k \cdot \frac{d}{dx}(f(x))$, we can rewrite this as:

$$\backslash cos\; 3x\; =\; \backslash frac\{d\}\{dx\}\; \backslash left(\; \backslash frac\{1\}\{3\}\; \backslash sin\; 3x\; \backslash right)$$

Thus, the anti-derivative of $\cos 3x$ is $\frac{1}{3}\sin 3x$.

Answer: The anti-derivative of $\cos 3x$ is $\backslash frac\{1\}\{3\}\backslash sin\; 3x$.

We need to find a function whose derivative is $e^{2x}$. We know that the derivative of $e^x$ is $e^x$. Let's consider the function $e^{2x}$.

The derivative of $e^{2x}$ with respect to $x$ is:

$\backslash frac\{d\}\{dx\}(e^\{2x\})\; =\; e^\{2x\}\; \backslash times\; \backslash frac\{d\}\{dx\}(2x)\; =\; 2e^\{2x\}$

We want to find a function whose derivative is $e^{2x}$. From the above, we can write:

$$e^\{2x\}\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{d\}\{dx\}\; (e^\{2x\})$$

Using the property of derivatives that $\frac{d}{dx}(k \cdot f(x)) = k \cdot \frac{d}{dx}(f(x))$, we can rewrite this as:

$$e^\{2x\}\; =\; \backslash frac\{d\}\{dx\}\; \backslash left(\; \backslash frac\{1\}\{2\}\; e^\{2x\}\; \backslash right)$$

Therefore, the anti-derivative of $e^{2x}$ is $\frac{1}{2}e^{2x}$.

Answer: The anti-derivative of $e^{2x}$ is $\backslash frac\{1\}\{2\}e^\{2x\}$.

We are looking for a function whose derivative is $(ax+b)^2$. We know that the derivative of $x^n$ is $nx^{n-1}$. Let's consider a function of the form $(ax+b)^3$.

The derivative of $(ax+b)^3$ with respect to $x$ is:

$$\backslash frac\{d\}\{dx\}((ax+b)^3)\; =\; 3(ax+b)^2\; \backslash times\; \backslash frac\{d\}\{dx\}(ax+b)\; =\; 3(ax+b)^2\; \backslash times\; a\; =\; 3a(ax+b)^2$$

We want to find a function whose derivative is $(ax+b)^2$. From the above, we can write:

$$(ax+b)^2\; =\; \backslash frac\{1\}\{3a\}\; \backslash frac\{d\}\{dx\}\; ((ax+b)^3)$$

Using the property of derivatives that $\frac{d}{dx}(k \cdot f(x)) = k \cdot \frac{d}{dx}(f(x))$, we can rewrite this as:

$$(ax+b)^2\; =\; \backslash frac\{d\}\{dx\}\; \backslash left(\; \backslash frac\{1\}\{3a\}\; (ax+b)^3\; \backslash right)$$

Thus, the anti-derivative of $(ax+b)^2$ is $\frac{1}{3a}(ax+b)^3$.

Answer: The anti-derivative of $(ax+b)^2$ is $\backslash frac\{1\}\{3a\}(ax+b)^3$.

We need to find a function whose derivative is $\sin 2x - 4e^{3x}$. We can find the anti-derivative of each term separately.

From Question 1, the anti-derivative of $\sin 2x$ is $-\frac{1}{2}\cos 2x$.

For the term $-4e^{3x}$, we first find the derivative of $e^{3x}$: $\frac{d}{dx}(e^{3x}) = 3e^{3x}$. So, $e^{3x} = \frac{1}{3}\frac{d}{dx}(e^{3x}) = \frac{d}{dx}(\frac{1}{3}e^{3x})$.

Therefore, the anti-derivative of $-4e^{3x}$ is $-4 \times \frac{1}{3}e^{3x} = -\frac{4}{3}e^{3x}$.

Combining the anti-derivatives of both terms, we get:

$\backslash frac\{d\}\{dx\}\backslash left(-\backslash frac\{1\}\{2\}\backslash cos\; 2x\; -\; \backslash frac\{4\}\{3\}e^\{3x\}\backslash right)\; =\; \backslash frac\{d\}\{dx\}\backslash left(-\backslash frac\{1\}\{2\}\backslash cos\; 2x\backslash right)\; -\; \backslash frac\{d\}\{dx\}\backslash left(-\backslash frac\{4\}\{3\}e^\{3x\}\backslash right)\; =\; \backslash sin\; 2x\; -\; 4e^\{3x\}$

Thus, the anti-derivative of $\sin 2x - 4e^{3x}$ is $-\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x}$.

Answer: The anti-derivative of $\sin 2x - 4e^{3x}$ is $\backslash left(-\backslash frac\{1\}\{2\}\backslash cos\; 2x\; -\; \backslash frac\{4\}\{3\}e^\{3x\}\backslash right)$.

To find the integral of $(4e^{3x} + 1)$, we can use the linearity property of integrals, which states that $\int (f(x) + g(x)) dx = \int f(x) dx + \int g(x) dx$ and $\int k f(x) dx = k \int f(x) dx$.

First, let's find the integral of $4e^{3x}$:

$\backslash int\; 4e^\{3x\}\; dx\; =\; 4\; \backslash int\; e^\{3x\}\; dx$

We know that the integral of $e^{ax}$ is $\frac{1}{a}e^{ax}$. So, for $a=3$, the integral of $e^{3x}$ is $\frac{1}{3}e^{3x}$.

$4\; \backslash int\; e^\{3x\}\; dx\; =\; 4\; \backslash left(\; \backslash frac\{1\}\{3\}e^\{3x\}\; \backslash right)\; =\; \backslash frac\{4\}\{3\}e^\{3x\}$

Next, let's find the integral of $1$:

$\backslash int\; 1\; dx\; =\; x$

Combining these results and adding the constant of integration $C$ (since it is an indefinite integral):

$$\backslash int\; (4e^\{3x\}\; +\; 1)\; dx\; =\; \backslash frac\{4\}\{3\}e^\{3x\}\; +\; x\; +\; C$$

Answer: The integral of $(4e^{3x} + 1)$ is $\backslash frac\{4\}\{3\}e^\{3x\}\; +\; x\; +\; C$.